# 概率公式总结

## 公式1

$\text{令}\mathrm T(x){为概率函数，}\mathrm G(x)\text{为在}x\text{点时的概率（一个变量），那么} \mathrm F(x) = \int_0^x\mathrm T‘(t)\mathrm G(t)\mathrm d t\text{为此时的概率函数。}(x \in [0,1])$

$\text{假定} F(1) = T(1) = 1.$

### 解答

$$\mathrm{T}_2(x) = 2x[x \in [0,0.5]]$$

$$\mathrm{T}_3(x)$$ = $$x\mathrm{T}_2(x) + \int_0^x \mathrm{T}'_2(t) (x-t) \mathrm d t (x \in [0,0.5])$$ \
= $$3x^2 (x \in [0,0.5])$$

$$\mathrm{T}_4(x)$$ = $$x\mathrm{T}_3(x) + \int_0^x \mathrm{T}'_3(t) (x - t) \mathrm d t (x \in [0,0.5])$$ \
= $$4x^3 (x \in [0,0.5])$$

Q. E. D.

Ex. 更特别地，观察$\mathrm{T}_2(x) = 2x,\mathrm{T}_3(x) = 3x^2,\mathrm{T}_0(x) = 4x^3$, 容易猜想 $$\mathrm{T}_n(x) = nx^{n-1}$$

$\mathrm{T}_2(x) = 2x^{2-1},$

$$\mathrm{T}_n(x) = x\mathrm{T}_{n-1}(x) + \int_0^x \mathrm{T}'_{n-1}(t) (x-t) \mathrm d t$$

$$\color{white}{\mathrm{T}_n(x)} = (n-1)x^{n-1} + \int_0^x (n-1)(n-2)t^{n-3}(x-t) \mathrm d t$$

$$\color{white}{\mathrm{T}_n(x)} = (n-1)x^{n-1} + (n-1)(n-2) \int_0^x xt^{n-3} -t^{n-2} \mathrm d t$$

$\color{white}{\mathrm{T}_n(x)}=n x ^{n-1}$